Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤10?5??) which is the total number of nodes, and a positive K (≤N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
解答：
该题可以用vector来模拟链表的实现，然后每k个一组保存在新的vector变量result中，最后输出结果。刚开始如果过于专注于节点顺序，会把题目搞复杂，要有大局观。
AC代码如下：
#include<iostream>
#include<vector>
#include<algorithm>#define maxn 1000000using namespace std;typedef struct{int data, nxt;
}node;typedef struct{int addr, data;
}node2;
vector<node> vec(maxn);int main()
{int addr, n, k;cin >> addr >> n >> k;for(int i = 0; i < n; ++i){int cur, d, nxt;scanf("%d %d %d", &cur, &d, &nxt);vec[cur].data = d;vec[cur].nxt = nxt;}vector<node2> list;for(int i = addr; i != -1; i = vec[i].nxt){node2 tmp;tmp.addr = i; tmp.data = vec[i].data;list.push_back(tmp);} vector<node2> result;int i;for(i = 0; i + k <= list.size(); i += k){int m = i, n = i + k - 1;while(n >= m){result.push_back(list[n]);n--;}}while( i < list.size() ){result.push_back(list[i++]);}for(int i = 0; i < result.size(); ++i){if(i != result.size() - 1) printf("%05d %d %05d\n", result[i].addr, result[i].data, result[i+1].addr);else printf("%05d %d -1\n", result[i].addr, result[i].data);}return 0;
}